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Equation (39) of [1] states the Baker-Campbell-Hausdorff formula for two operators \( a, b\) that commute with their commutator \( \antisymmetric{a}{b} \)

\begin{equation}\label{eqn:bakercambell:20}

e^a e^b = e^{a + b + \antisymmetric{a}{b}/2},

\end{equation}

and provides the outline of an interesting method of proof. That method is to consider the derivative of

\begin{equation}\label{eqn:bakercambell:40}

f(\lambda) = e^{\lambda a} e^{\lambda b} e^{-\lambda (a + b)},

\end{equation}

That derivative is

\begin{equation}\label{eqn:bakercambell:60}

\begin{aligned}

\frac{df}{d\lambda}

&=

e^{\lambda a} a e^{\lambda b} e^{-\lambda (a + b)}

+

e^{\lambda a} b e^{\lambda b} e^{-\lambda (a + b)}

–

e^{\lambda a} b e^{\lambda b} (a + b)e^{-\lambda (a + b)} \\

&=

e^{\lambda a} \lr{

a e^{\lambda b}

+

b e^{\lambda b}

–

e^{\lambda b} (a+b)

}

e^{-\lambda (a + b)} \\

&=

e^{\lambda a} \lr{

\antisymmetric{a}{e^{\lambda b}}

+

{\antisymmetric{b}{e^{\lambda b}}}

}

e^{-\lambda (a + b)} \\

&=

e^{\lambda a}

\antisymmetric{a}{e^{\lambda b}}

e^{-\lambda (a + b)}

.

\end{aligned}

\end{equation}

The commutator above is proportional to \( \antisymmetric{a}{b} \)

\begin{equation}\label{eqn:bakercambell:80}

\begin{aligned}

\antisymmetric{a}{e^{\lambda b}}

&=

\sum_{k=0}^\infty \frac{\lambda^k}{k!} \antisymmetric{a}{ b^k } \\

&=

\sum_{k=0}^\infty \frac{\lambda^k}{k!} k b^{k-1} \antisymmetric{a}{b} \\

&=

\lambda \sum_{k=1}^\infty \frac{\lambda^{k-1}}{(k-1)!} b^{k-1}

\antisymmetric{a}{b} \\

&=

\lambda e^{\lambda b} \antisymmetric{a}{b},

\end{aligned}

\end{equation}

so

\begin{equation}\label{eqn:bakercambell:100}

\frac{df}{d\lambda} = \lambda \antisymmetric{a}{b} f.

\end{equation}

To get the above, we should also do the induction demonstration for \( \antisymmetric{a}{ b^k } = k b^{k-1} \antisymmetric{a}{b} \).

This clearly holds for \( k = 0,1 \). For any other \( k \) we have

\begin{equation}\label{eqn:bakercambell:120}

\begin{aligned}

\antisymmetric{a}{b^{k+1}}

&=

a b^{k+1} – b^{k+1} a \\

&=

\lr{ \antisymmetric{a}{b^{k}} + b^k a

} b – b^{k+1} a \\

&=

k b^{k-1} \antisymmetric{a}{b} b

+ b^k \lr{ \antisymmetric{a}{b} + {b a} }

– {b^{k+1} a} \\

&=

k b^{k} \antisymmetric{a}{b}

+ b^k \antisymmetric{a}{b} \\

&=

(k+1) b^k \antisymmetric{a}{b}.

\end{aligned}

\end{equation}

Observe that \ref{eqn:bakercambell:100} is solved by

\begin{equation}\label{eqn:bakercambell:140}

f = e^{\lambda^2\antisymmetric{a}{b}/2},

\end{equation}

which gives

\begin{equation}\label{eqn:bakercambell:160}

e^{\lambda^2 \antisymmetric{a}{b}/2} =

e^{\lambda a} e^{\lambda b} e^{-\lambda (a + b)}.

\end{equation}

Right multiplication by \( e^{\lambda (a + b)} \) which commutes with \( e^{\lambda^2 \antisymmetric{a}{b}/2} \) and setting \( \lambda = 1 \) recovers \ref{eqn:bakercambell:20} as desired.

What I wonder looking at this, is what thought process led to trying this in the first place? This is not what I would consider an obvious approach to demonstrating this identity.

# References

[1] Roy J Glauber. Some notes on multiple-boson processes. *Physical Review*, 84 (3), 1951.